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Integral calculus is a fundamental branch of calculus that focuses on finding the integral of functions. At its core, it seeks to determine the accumulation of quantities, often represented as an area under a curve. The process involves reversing differentiation or finding the antiderivative. Integrals are crucial in mathematical analysis for calculating areas, volumes, and other accumulated values.
When we talk about integrals in this context, it's usually about definite and indefinite integrals. An indefinite integral, like the one we're solving here, includes a constant of integration, typically denoted by \( C \), because integrating a function involves an infinite set of antiderivatives.
Integration by parts is a specific technique within integral calculus, useful when integrating the product of two functions. The integration by parts formula is derived from the product rule for differentiation, and it's expressed as:

• \( \int u \, dv = uv - \int v \, du \)

This formula cleverly shifts the integration problem into a potentially simpler form, usually requiring strategic choices for \( u \) and \( dv \) to streamline the process.

The natural logarithm function, denoted as \( \ln(t) \), is a logarithm to the base \( e \), where \( e \) is approximately 2.71828. It is an essential mathematical function, especially in calculus, due to its continuous, increasing nature and its properties that simplify complex calculations.
In integration, \( \ln(t) \) often appears in expressions where its derivatives and integrals interact smoothly with algebraic terms. The derivative of the natural logarithm function \( \ln(t) \) is particularly important because it simplifies as:

• \( \frac{d}{dt}\ln(t) = \frac{1}{t} \)

Using this derivative simplifies part of the integration by parts process, as seen in the solution's step-by-step breakdown. Recognizing this, we select \( u = \ln(t) \) in the given exercise to leverage the simplicity of its derivative, \( du = \frac{1}{t}\,dt \). This choice directly supports the computational steps for integration by parts.

Working through problems with a step-by-step solution is a powerful way to learn complex mathematical techniques, such as integration by parts. This methodical approach helps clarify each piece of the problem, making it easier to follow and understand.
In this exercise, we first identify key components \( u \) and \( dv \) to simplify the integral. By choosing \( u = \ln(t) \) and \( dv = \frac{1}{t^2} \, dt \), we set the stage for a easier calculation. Differentiating \( u \) yields \( du = \frac{1}{t} \, dt \), and integrating \( dv \) provides \( v = -\frac{1}{t} \). These components are pivotal for integration by parts.
Once you've assigned \( u \) and \( dv \), substitute into the integration by parts formula:

• \( \int \frac{\ln t}{t^2} \, dt = \ln t \left(-\frac{1}{t}\right) - \int \left(-\frac{1}{t}\right) \frac{1}{t} \, dt \)

Rewriting and simplifying it, we finally resolve the integral to:

• \(-\frac{\ln t}{t} - \frac{1}{t} + C \)

This step-by-step analysis ensures that you understand each transformation, allowing you to handle similar problems in the future with confidence.